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jeka57 [31]
3 years ago
10

An object is accelerating if it is moving____. circle all that apply. ​

Physics
1 answer:
tino4ka555 [31]3 years ago
3 0

Answer:

With changing speed and/or in a circle

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At what angle two forces P + Q and (P - Q) act so that their resultant is :
stiv31 [10]

Use resultant formula

\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}

So

#1

A be p+q and B be p-q

\\ \rm\Rrightarrow R=\sqrt{3p^2+q^2}

\\ \rm\Rrightarrow \sqrt{(p+q)^2+(p-q)^2+2(p+q)(p-q)cos\alpha}=\sqrt{3p^2+q^2}

\\ \rm\Rrightarrow 2p^2+2q^2+2(p^2-q^2)cos\alpha=3p^2+q^2

\\ \rm\Rrightarrow 2cos\alpha=1

\\ \rm\Rrightarrow cos\alpha=\dfrac{1}{2}

\\ \rm\Rrightarrow \alpha=\dfrac{\pi}{3}

#2

\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\beta=2(p^2+q^2)

\\ \rm\Rrightarrow 2cos\beta=0

\\ \rm\Rrightarrow cos\beta=0

\\ \rm\Rrightarrow \beta=\dfrac{\pi}{2}

#3

\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\gamma=p^2+q^2

\\ \rm\Rrightarrow 2(p^2-q^2)cos\gamma=-(p^2+q^2)

\\ \rm\Rrightarrow cos\gamma =\dfrac{q^2-p^2}{2(p^2-q^2)}

\\ \rm\Rrightarrow \gamma=cos^{-1}\left(\dfrac{q^2-p^2}{2(p^2+q^2)}\right)

8 0
2 years ago
Read 2 more answers
Solve the inequality. x/3 is greater than or equal to - 6. a. x ≥ –9 b. x ≥ 9 c. x ≥ –18 d. x ≤ –18
NISA [10]
To get x on its own, you times the 3 over to the other side so the 3 cancels out on the LHS. 

~ x greater than or equal to -18

(C)
6 0
3 years ago
A 25 kg child runs at a speed of 5.0 m/s and jumps onto a stationary shopping cart and holds on for dear life. The cart has mass
makkiz [27]

Answer:

3.38 m/s

Explanation:

Mass of child = m₁ = 25

Initial speed of child = u₁ = 5 m/s

Initial speed of cart = u₂ = 0 m/s

Mass of cart = m₂ = 12 kg

Velocity of cart with child on top = v

This is a case of perfectly inelastic collision

m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\frac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\frac{25\times 5+12\times 0}{25+12}\\\Rightarrow v=\frac{125}{37}\\\Rightarrow v=3.38\ m/s

Velocity of cart with child on top is 3.38 m/s

7 0
3 years ago
Select the option that best summarizes the reason for the use of position-time graphs when studying physical science. (1 point)
Sedaia [141]

The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.

<h3>What is velocity?</h3>

Velocity is a vector quantity that tells the distance an object has traveled over a period of time.

Displacement is a vector quality showing total length of an area traveled by a particular object.

Imagine a time-position graph where the velocity of an object is constant. What will be observed on the graph concerning the slope of the line segment as well as the velocity of the object?

The slope of the line is equal to zero and the object will be stationary.

The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.

To learn more about velocity refer to the link

brainly.com/question/18084516  

#SPJ2

4 0
1 year ago
Read 2 more answers
A bullet of mass 11 g strikes a ballistic pendulum of mass 1.9 kg. The center of mass of the pendulum rises a vertical distance
Ymorist [56]

Answer:

The bullet's initial speed is 243.21 m/s.

Explanation:

Given that,

Mass of the bullet, m_b=11\ g=0.011\ kg

Mass of the pendulum, m_p=19\ kg

The center of mass of the pendulum rises a vertical distance of 10 cm.

We need to find the bullet's initial speed if it is assumed that the bullet remains embedded in the pendulum. Let it is v. In this case, the energy of the system remains conserved. The kinetic energy of the bullet gets converted to potential energy for the whole system. So,

\dfrac{1}{2}(m_b+m_p)V^2 =(m_b+m_p)gh\\\\V=\sqrt{2gh} \ .................(1)

V is the speed of the bullet and pendulum at the time of collision

Now using conservation of momentum as :

m_bv=(m_p+m_b)V

Put the value of V from equation (1) in above equation as :

v=\dfrac{(m_p+m_b)}{m_b}\sqrt{2gh} \\\\v=\dfrac{(1.9+0.011)}{0.011}\sqrt{2\times 9.8\times 0.1}\\\\v=243.21\ m/s

So, the bullet's initial speed is 243.21 m/s.

7 0
3 years ago
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